HW9-10Sol

# HW9-10Sol - Oregon State University Physics 201 Fall 2009...

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Unformatted text preview: Oregon State University Physics 201 , Fall 2009 HW9 10 (due Dec. 4 at 5:00 p.m.) Page 1 Oregon State University Physics 201 Fall Term, 2009 HW9-10 Solutions 1. A 710-kg car drives at a constant speed of 23 m/s. It is subject to a drag force of 500 N. What power is required from the carʼs engine to drive the car: (a) on level ground? (b) up a hill with a slope of 2.0°? (a) In order to maintain a constant speed, the road must push on the car (i.e. the carʼs tires must push on the road) with a forward force equal but opposite to the drag force: F forward = F D = 500 N And the power supplied via this force is easily calculated: P forward = F forward v forward Doing the numbers: P forward = (500)(23) = 1.15 x 10 4 W (b) To allow the car to climb the hill at that same constant speed, now the forward force must balance the sum of the drag force and the downhill component of the gravitational force: F forward = F D + mg sin θ = 500 + (710)(9.80)·sin2° = 742.83 And again, the power supplied via this force is easily calculated: P forward = F forward v forward Doing the numbers: P forward = (742.83)(23) = 1.71 x 10 4 W Oregon State University Physics 201 , Fall 2009 HW9 10 (due Dec. 4 at 5:00 p.m.) Page 2 2. A certain roller-coaster in California has a circular “loop-the-loop” feature with a radius of 7.00 m. It is designed so that riders are in “free-fall” for the moment when they are at the top of the loop. What speed must the roller coaster have as it enters the loop and begins to climb? (Assume no friction in the track and ignore the slow rotation of the roller coaster.) Analyze the situation from an initial point at the bottom of the loop, where the roller coaster is just entering it (at speed v i ), to the top of the loop, where the roller coaster is moving at (tangential) speed v f . f For the purposes of measuring U G , let h = 0 at the bottom of the loop. The fact that the roller coaster is in free-fall at the top indicates that its weight is the only y-force (i.e. the normal force by the track is momentarily zero), so its weight is equal to the net y-force— the centripetal force that keeps the roller coaster moving in a circular path: mg = mv f 2 / r Simplifying this, we have g = v f 2 / r . Solving for v f 2 ( v f 2 = rg ), we can substitute it into the above result: v i 2 = rg + 4 rg Collect terms: v i 2 = 5 rg Take the square root: v i = (5 rg ) 1/2 The numbers: v i = [5(7.00)(9.80)] 1/2 = 18.5 m/s 7.00 m Analysis of the work being done: There is no work being done by any force except gravity (so W ext = 0). There is the normal force by the track on the roller coaster—but it is always at right angles to the displacement and hence contributes no work....
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HW9-10Sol - Oregon State University Physics 201 Fall 2009...

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